PROP. XVII. PROB. From a given point( A ) without a given circle ( CBF), Fig. 24. to draw a right line which shall be a tangent to the circle. Let C be the centre of the given circle, and from the centre C with the radius CA, describe a circle CAE; draw CA which meets the circle in the point F, and draw through the point F the line FE perpendicular to CA and meeting the circle CAE in E, draw the line CE meeting in B the given circle, and the right line drawn from B to the given point A is a tangent. For in the triangles ACB, ECF, the sides AC and CB are equal to EČ and CF (1), and the angle at C is (1)Def. 15 common to both, therefore the angle ABC is equal to EFC (2), but the angle EFC is a right angle (3), there- (2) Prop.4. fore ABC is a right angle, and therefore the right line B. 1. AB is a tangent to the circle CFB (4). (3) Constr. Schol. It is evident that there can be drawn two (4) Prop.16 tangents from the point A, one at either side of the B. 3. right line AC. B. I. PROP. XVIII. THEOR. If a right line (DB) be a tangent to a circle, the Fig. 25. right line (CD) drawn from the centre to the point of contact, is perpendicular to it. For, if it be possible, let the right line CF be perpendicular to BD, and in the triangle CFD, because the angle CFD is a right angle, the angle CDF is acute (1), therefore the side CD is greater than the side (1) Cor. CF (2), but CE is equal to CD, and therefore CE is 1.17.8. 1. greater than CF, a part greater than the whole, which (2)Prop. 19 is absurd. Therefore CF is not perpendicular to BD, B. 1. and in the same manner it can be demonstrated that no other line except CD is perpendicular to it. K PROP. XIX. THEOR. Fig. 26. If a right line (BC) be a tangent to a circle, the right line (BA) drawn perpendicular to it from the point of contact, passes through the centre of the circle. For, if it be possible, let the centre Z be without the line BA and draw ZB. Because the right line ZB is drawn from the centre to the point of contact, it is perpendicular to the tan(1)Prop.18 gent (1), therefore the angle ZBC is a right angle, but the angle ABC is also a right angle (2), and therefore (2) Hypoth. ZBC is equal to ABC, a part to the whole, which is absurd. Therefore Z is not the centre, and in the same manner it can be demonstrated, that no other point without the line AB is the centre. B. 3. PROP. XX. THEOR. Fig. 27. The angle (ACD) at the centre of a circle is double 28. & 29. of the angle (ABD) at the circumference, when they See N. have the same part of the circumference for their base. Fig. 27. 1. Let one side of the angle at the circumference pass through the centre; because in the triangle DCB, the sides DC and CB are equal, the angles CBD and CDB (1) Prop. 5. are equal (1), but the external angle ACD is equal to B. 1. both of them taken together, (2), and therefore is dou(2)Prop.32 ble of ABD. B. 1. Fig. 28. 2. Let the angle ACD fall within the angle ABD, and draw through the centre the right line BCE; be(3) First cause the angle ACE is double of the angle ABC (3), pürt. and the angle DCE is double of the angle DBC (3), the whole angle ACD is equal to double the angle ABC and double DBC, and therefore equal to double the angle ABD. Fig. 29, 3. Let one side of the angle ACD cut a side of the angle ABD, and draw through the centre the right line BCE; the angle ECD is equal to double the angle EBD (4), or to double the angle EBA together with (4) First double the angle ABD, but the angle ECA is equal to part. double the angle EBA (4), take away these equal quantities from both, and the angle ACD shall be equal to double the angle ABD. PROP. XXI. THEOR. The angles (BAD, BED) in the same segment of a Fig. 30 & circle are equal. 51. B. 3. 1. Let the segment BAD be greater than a semi- Fig. 30. circle, let C be the centre of the circle, and draw CB and CD, The angle BCD at the centre is double of the angle (1) Prop.20 BAD, and also double of BED (1), therefore BAD B: 3: and BED are equal to one another (2). (2) Ax. 7. 2. Let the segment BAD be a semicircle or less Fig. 31. than a semicircle, let C be the centre of the circle, and draw the right lines ACF and EF. Because the segment BDF is greater than a semicircle, and in it are the angles BAF and BEF, BAF is equal to BEF (3); and because the segment FBAD (3) First is greater than a semicircle, and in it are the angles part. FAD and FED, FAD is equal to FED (3), therefore the sum of the angles BAF and FAD, or the angle BAD, is equal to the sum of BEF and FED, or to the angle BED. Cor. If two equal angles stand upon the same arch Fig. 32. BD and the vertex of one of them BAD be in the circumference of the circle, the vertex of the other shall be in the same circumference, For, if it be possible, let the vertex of the other angle fall either without or within the circumference, as at F, and draw FD; because the angles BAD and BED are in the same segment, they are equal (1), but (1) Prop.21 BAD is equal to BFD (2), and therefore BED is (2) Hypoth. equal to BFD, but one of them is greater than the (3) Prop.16 other (3), which is absurd: therefore the point FB. 1. neither falls within nor without the circle, and therefore it falls upon the circumference itself. B. 3. PROP. XXII. THEOR. Fig. 33. The opposite angles of a quadrilateral figure(FABC) inscribed in a circle, are together equal to two right angles. Draw the diagonals AC and FB. Because the angles ACB and AFB are in the same (1) Prop. segment AFCB, ACB is equal to AFB (1); and be21. B. 3. cause the angles ACF and ABF are in the same seg ment ABCF, ACF is equal to ABF (1); therefore the angle BCF is equal to the angles AFB and ABF taken together, but the angles AFB and ABF together with (2)Prop.32 FAB are equal to two right angles (2), and therefore BCF together with FAB is equal to two right angles: in the same manner it can be demonstrated that ABC and AFC are equal to two right angles. Cor. If one of the sides of a quadrilateral figure in scribed in a circle be produced, the external angle is (1) Prop. prec. & equal to the internal remote angle, for each of them Prop. 13. together with the internal adjacent angle is equal to B. }, two right angles (1). B. l. a PROP. XXIII. THEOR. Fig. 34, Upon the same right line, and upon the same side of it, tuo similar segments of circles cannot be constructed which do not coincide. For, if it be possible, let two similar segments ACB and ADB be constructed, and let the point D in one of them fall without the other, and draw the right lines DA, DB and CB. Because the segments ACB and ADB are similar, (1) Def.10. the angle ACB is equal to ADB (1), but ACB is ex& Prop. 21. ternal to ADB and therefore greater than it (2), which . B. 3. is absurd: therefore no point in either of the segments (2) Prop.16 falls without the other, and therefore the segments coincide. B. 3. PROP. XXIV. THEOR. Similar segments of circles, standing upon equal right Fig. 35. lines ( AB and CD), are equal. For if the equal right lines AB and CD be so applied to one another, that the point A may fall on C, the point B must fall upon D and therefore the right (2) Prop.28 (1) At. 10 ( lines coincide (1), therefore the segments themselves B. 3. coincide(2), and therefore are equal. PROP. XXV. PROB. A segment (ABC) of a circle being given, to describe Fig. 36. the circle of which it is the segment. From any point B draw two right lines BA and BC, bisect them, and from the points of bisection F and E draw two lines FO and EO, perpendicular to AB and BC, the intersection 0 of these perpendiculars is the centre, Because the right line AB terminated in the circle is bisected by a perpendicular to it FO, FO passes through the centre (i), likewise EO passes through (1) Schol. the centre (1), therefore the centre must be in 0 the Prop. l. intersection of these lines FO and EO. See N. B. 3. PROP. XXVI, THEOR, In equal circles (ABC, DEF) equal angles (AOC Fig. 37. See N. and DĀF, ABC and DEF),whether they be at the centres or at the circumferences, stand upon equal arches, First let the given angles AOC and DHF be at the centres, draw to any points B and E in the circumferences the lines AB, CB, and DE, FE, and join AC and DF. Because in the triangles AOC, DHF, the angles O (1) hypoth. and H are equal (1), and the sides AO and OC equal (2) Prop.4. to DH and HF (1), the bases AC and DF are equal (3)Prop. 20 (2); but the angles ABC and DEF are equal (3), and far. 7. therefore the segments ABC and DEF are similar (4), (4)Def. 16. but they stand upon equal right lines AC and DF, and B.3. are therefore equal(5); take awaythese equals from the (5) Prop.24 B. 1. 3 |